Question

A long jumper leaves the ground with an initial velocity of 10m/s at an angles of 27° above the horizontal. What is the time of flight, maximum height, and range?

Answer 1

To solve the problem, we can break down the initial velocity into horizontal and vertical components.

Given: Initial velocity (v0) = 10 m/s Launch angle (θ) = 27°

1. Time of Flight: The time of flight (T) is the total duration of the jump. We can calculate it using the vertical component of velocity.

The vertical component of velocity (v0y) can be calculated as: v0y = v0 * sin(θ)

T = (2 * v0y) / g where g is the acceleration due to gravity (approximately 9.8 m/s²).

Substituting the values: v0y = 10 m/s * sin(27°) T = (2 * v0y) / g

2. Maximum Height: The maximum height (H) is the highest point reached by the jumper. We can calculate it using the vertical component of velocity and the equation for vertical motion.

The equation for vertical motion is: H = (v0y²) / (2 * g)

Substituting the values: H = (v0y²) / (2 * g)

3. Range: The range (R) is the horizontal distance covered by the jumper. We can calculate it using the horizontal component of velocity and the equation for horizontal motion.

The horizontal component of velocity (v0x) can be calculated as: v0x = v0 * cos(θ)

The equation for horizontal motion is: R = v0x * T

Substituting the values: v0x = 10 m/s * cos(27°) R = v0x * T

Now we can calculate the values:

First, calculate v0y, v0x, and T: v0y = 10 m/s * sin(27°) v0x = 10 m/s * cos(27°) T = (2 * v0y) / g

Next, calculate H and R: H = (v0y²) / (2 * g) R = v0x * T

By substituting the values and performing the calculations, you can find the specific numerical values for time of flight, maximum height, and range.